Double integral in Rectangular coordinates for anything circular

  • Thread starter PaultheRiemann
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  • #1
PaultheRiemann
This is the equation for the cone
A \sqrt{x^2 + y^2}

The double integral

\iint A \sqrt{x^2 + y^2} \space dy \space dx \space \space \space


\text {From x= -1 to 1 and y=} -\sqrt{1-x^2} \space to \space \sqrt{1-x^2}

\text{ is very difficult to evaluate. I've tried polar coordinate substitution. However, I can't seem to get} /space 2/3 \pi z \space \text{where z is the height of cone and 1 is the radius of the cone}
 

Answers and Replies

  • #2
PaultheRiemann
\text{here's what happens when I use CLAIRAUT theorem to change the order of the pertaining triple integral}

ImageUploadedByPhysics Forums1420297046.972458.jpg
 
  • #3
PaultheRiemann
I'm also trying my best not to use cylindrical and spherical coordinate systems.
 
  • #4
PaultheRiemann
ImageUploadedByPhysics Forums1420298926.716545.jpg
easy when spherical coordinates are used
 
  • #5
HallsofIvy
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You are not getting any responses, perhaps because no one can understand what you are saying.
Put your Latex between double dollar signs.
This is the equation for the cone
$$A \sqrt{x^2 + y^2}$$
No, it isn't, in fact, it is not even an equation! I assume you meant $$z= A\sqrt{x^2+ y^2}$$ but even that is only one side of the cone- the entire cone is $$z^2= A^2(x^2+ y^2)$$

The double integral

$$\iint A \sqrt{x^2 + y^2} \space dy \space dx \space \space \space$$


$$\text {From x= -1 to 1 and y=} -\sqrt{1-x^2} \space to \space \sqrt{1-x^2}$$

\text{ is very difficult to evaluate. I've tried polar coordinate substitution. However, I can't seem to get} /space 2/3 \pi z \space \text{where z is the height of cone and 1 is the radius of the cone}
To integrate $\int_{x= -1}^1\int_{y= {-\sqrt{1- x^2}}^\sqrt{1- x^2} A\sqrt{x^2+ y^2}dy dx$$
I would first use trig identity $$tan^2(\theta)+ 1= sec^2(\theta)$$ and make the substitution $$y= xtan(\theta)$$. Then $$\sqrt{x^2+ y^2}= \sqrt{x^2+ x^2 tan(\theta)}= x sec(\theta)$$. Also $$dy= sec^2(\theta)d\theta$$
This is the equation for the cone
A \sqrt{x^2 + y^2}

The double integral

\iint A \sqrt{x^2 + y^2} \space dy \space dx \space \space \space


\text {From x= -1 to 1 and y=} -\sqrt{1-x^2} \space to \space \sqrt{1-x^2}

\text{ is very difficult to evaluate. I've tried polar coordinate substitution. However, I can't seem to get} /space 2/3 \pi z \space \text{where z is the height of cone and 1 is the radius of the cone}
No, it is not that difficult. The substitution $$y= x tan(\theta)$$ works nicely.

But I don't see why you are doing that. if you want to find the volume of a cone, with height h, then the double integral you are using isn't going to do it. Instead, note that, for any z, a cross section is a disk with center (0, 0, z) and radius $$r= \sqrt{x^2+ y^2}= \frac{z}{A}$$ so area $$\pi \frac{z^2}{A^2}$$. Taking the "thickness" to be dz, the volume of such disk is $$\pi \frac{z^2}{A^2}dz$$. The entire cone, thought of as the sum of such disks is $$\frac{\pi}{A^2}\int_0^h z^2 dz$$. In this particular cone, with $$z= A\sqrt{x^2+ y^2}$$ a cone with height h will have base radius $$\frac{h}{A}$$.

That will NOT give "$$\frac{2}{3}\pi h$$" because that is NOT the formula for volume of a cone. The volume of a cone, of height h and base radius R, is $$\frac{1}{3}\pi r^2h$$- 1/3, not 2/3.
 
  • #6
PaultheRiemann
Thanks halls
 

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